# Convergence of Taylor Series of Function Analytic on Disk/Lemma

## Lemma

Let $y > 1$.

Then:

$\displaystyle \lim_{n \mathop \to \infty} \frac n {y^n} = 0$

## Proof

Note that $\ln y > 0$ as $y > 1$.

 $\displaystyle \lim_{n \mathop \to \infty} \frac n {y^n}$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \frac n {\left({e^{\ln y} }\right)^n}$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \frac n {e^{\left({\ln y}\right) n} }$ $\displaystyle$ $=$ $\displaystyle 0$ as $\displaystyle \lim_{x \mathop \to \infty} \frac x {e^{\left({\ln y}\right) x} } = 0$ by Limit at Infinity of Polynomial over Complex Exponential as $\ln y > 0$

$\blacksquare$